Domain of Ln (1 Viewer)

[icycloud]

When asked to find the domain of something like:

ln(x+3) - ln(x-2)

(and hence to sketch the curve of)

Do we first simplify the expression into ln([x+3]/[x-2]) ?

If left unsimplified, one would think the domain is x > 2

However, once simplified we get x>2 or x<-3

Take for example, x = -5

If unsimplified, we get the expression ln(-2) - ln(-7) [you can't take the logs of negative numbers blah blah]

If simplified, we get ln(-2/-7) = ln(2/7) which is perfectly legitimate.

Perhaps it has to do with the definition e^(i*pi) = -1, thus ln(-1) = i*pi

If we use that definition,

x = -5,
ln(-5+3) - ln(-5-2) = ln(-2) - ln(-7)
= i*pi + ln(2) - i*pi - ln(7)
= ln(2) + ln(7)
= ln(2/7) as required

Hmm... Any thoughts? I probably missed something really simple.

 

[Raginsheep]

Just look at the two bits seperately.
For ln(x+3), x>-3. For ln(x-2), x>2.
Thus the domain is the region which satsifies all conditions, x>2.

As for your use of euler's formula, Im not sure if thats correct but it's not required in the 4u course.

 

[icycloud]

Just look at the two bits seperately.
For ln(x+3), x>-3. For ln(x-2), x>2.
Thus the domain is the region which satsifies all conditions, x>2.

As for your use of euler's formula, Im not sure if thats correct but it's not required in the 4u course.

That doesn't really answer my question.

 

[Raginsheep]

You asked when finding the domain of functions, do you simply it or work it out from the unsimplied form. As far as I can tell, I told you to use the unsimplified form. If I misinterpreted your question, then pardon me.

 

[KFunk]

I would do it by considering f(x) = (x+3)/(x-2) = 1 + 5/(x - 2)

This curve has a horizontal asymptote at y = 1 and a vertical one at x = 2 as well as a zero at x = -3. All you really need to know is where f(x) is &le; 0. Using simple graph drawing or otherwise you can determine that f(x) &le; 0 when -3 &le; x < 2 giving you the domain x>2 or x<3 for g(x) = ln(x+3) - ln(x-2)

 

[icycloud]

You asked when finding the domain of functions, do you simply it or work it out from the unsimplied form. As far as I can tell, I told you to use the unsimplified form. If I misinterpreted your question, then pardon me.

Yes, that was my question. But you did not explain why to use the unsimplified form as opposed to simplifying it? *confused* Sorry for my ignorance.

p.s. I plotted the function on Mathematica and its domain extends to x<-3 as well as x>2.

 

[Raginsheep]

I think the reason may be because simplying it may casue you to lose information regarding the nature of the curve.

For example, if you have y=x^2/x, the domain is all real x, x=/=0. If you simplified it, you would get y=x which is similar but doesn't have an restriction on its domain.

 

[Stan..]

I would do as Kfunk has done. To sketch them, sketch each one individually and them add the coodinates. Then use calculus to find the critical points.